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Question: A conductor with rectangular cross section has dimensions \(\left( {4a \times 2a \times a} \right)\)...

A conductor with rectangular cross section has dimensions (4a×2a×a)\left( {4a \times 2a \times a} \right) as shown in figure. Resistance across ABAB is xx, across CDCD is yy and across EFEF is ZZ.

(A) x=y=zx = y = z
(B) x>y>zx > y > z
(C) y>x>zy > x > z
(D) x>z>yx > z > y

Explanation

Solution

Useful formula: Use the relation between the resistance and resistivity to calculate the value of resistance at the point x,yx,\,y and zz. Substitute the values of length and cross sectional area given in the formula. Compare the values of obtained resistance of x,yx,\,y and zzto find the answer for this question.

Formulae Used:
Resistance formula is
R=ρlAR = \dfrac{{\rho l}}{A}
Where RR is the resistance of the conductor, ρ\rho is the resistivity of the conductor, ll is the length of the conductor and AA is the cross sectional area of the conductor.

Complete step-by-step solution:
The given data from the question are
Resistance across ABAB is xx
Resistance across CDCD is yy
Resistance across EFEF is zz
By using the resistance formula,
The resistance across xx is R=ρlAR = \dfrac{{\rho l}}{A}. Cross sectional area at xx is 4a×a4a \times a.
Substituting the values at the above equation
x=ρ(4a)2a×ax = \dfrac{{\rho \left( {4a} \right)}}{{2a \times a}}
x=ρ(2a)a2x = \dfrac{{\rho \left( {2a} \right)}}{{{a^2}}}
By simplifying the above equation,
x=2ρax = \dfrac{{2\rho }}{a}
Cross sectional area at yy is 4a×2a4a \times 2a.
Similarly, finding the values of resistance at yy.
y=ρ8ay = \dfrac{\rho }{{8a}}
Cross sectional area at zz is 4a×a4a \times a.
In the same way, finding the values of resistance at zz.
z=ρ2az = \dfrac{\rho }{{2a}}
From the obtained values of x,yx,\,y and zz, it is known that x>z>yx > z > y.
Thus the option (D) is correct.

Note:- Remember that the cross sectional area at a particular point is calculated by the sectional area which is obtained by a sliced section of the considered object. For example: the cross sectional area at xx is calculated as the area of the cross section of AB which is 2a×a2a \times a.